Section 1 Quiz 1-A Convert the following numbers from the given base to the other three bases listed in the table: Decimal Binary Hexadecimal 1377.140625 10101100001.001001 561.24 454.3125 111000110.0101 1C6.5 Simplify the following Boolean expressions to expressions containing a minimum number of literals Name: Number:
Section 2 Quiz 1-A Convert the following numbers from one notation to another. 42.375 10 = 101010.011 2 11101101000001 2 = 3B41 16 11001.101 2 = 25.625 10 AC9 16 = 101011001001 2 For each expression below, use DeMorgan s theorem to obtain an equivalent expression which contains ANDs and ORs of the inputs (e.g. A) and their complements (e.g. A ). There should be no complements (bars) in the final expression except those over the inputs. Name: Number:
Section 1 Quiz 1-B Convert the following numbers from the given base to the other three bases listed in the table: Decimal Binary Hexadecimal 369.3125 F3C7.A Simplify the following Boolean expressions to expressions containing a minumium number of literals (a) A B C + A C (b) A B D + A C D + B D Name: Student No.: -
Section 1 Quiz 1-B Convert the following numbers from the given base to the other three bases listed in the table: Decimal Binary Hexadecimal 369.3125 101110001.0101 171.5 62407.625 1111001111000111.101 F3C7.A Simplify the following Boolean expressions to expressions containing a minimum number of literals (a) (b) Name: Student No.: -
Section 2 Quiz 1-B Convert the following numbers from one notation to another. 25.625 10 = 11001.101 2 1101111010101111 2 = DEAF 16 10101.111 2 = 21.875 10 1ED5 16 = 1111011010101 2 For each expression below, use DeMorgan s theorem to obtain an equivalent expression which contains ANDs and ORs of the inputs (e.g. A) and their complements (e.g. A ). There should be no complements (bars) in the final expression except those over the inputs. Name: Number:
4 0907231 Digital Logic () Quiz 1 Form A Solutions are in RED COLOR رقم التسجيل : الاسم : الشعبة: رقم ====================================================================== Instructions: Time 15 minutes. Closed books and notes. No calculators. No questions are allowed. ======================================================================= Q1. Convert (325) 10 to binary. (8 points) 325 162 1 162 81 0 81 40 1 40 20 0 20 10 0 10 5 0 5 2 1 2 1 0 1 0 1 (325) 10 = 101000101 OR: 325-256 = 69 69-64 = 5 5 4 = 1 (325) 10 = 256+64+4 + 1 = 2 8 + 2 6 + 2 2 + 2 0 = 101000101 Q2. Convert (A63) 16 to binary. (4 points) (A63) 16 = 1010 0110 0011 Q3. Analyze the following circuit to find Boolean Functions F(X,Y,Z) are being implemented. Then simplify your answer if possible. (16 points) F(X,Y,Z) = X YZ +X YZ + XZ Before simplification = X Y(Z +Z ) + XZ By identity 14 = X Y.1 + XZ By identity 7 = X Y + XZ By identity 2 F(X,Y,Z) = X Y + XZ After simplification 1
Q4. Evaluate the following binary addition. (4 points) carry 1 1 1 1 0 1 1 1 0 Q5. From the truth table, write the function F(X,Y,Z) in the SOP terms then draw circuit implementing the followin function using AND, Or, and NOT gates without simplification. (18 points). (8 points for the function and 10 points for the circuit) X Y Z F minterm 0 0 0 1 X Y Z 0 0 1 1 X Y Z 0 1 0 0 0 1 1 0 1 0 0 1 XY Z 1 0 1 0 1 1 0 1 XYZ 1 1 1 1 XYZ F(X,Y,Z) = X Y Z + X Y Z + XY Z + XYZ + XYZ 2
3 4 0907231 Digital Logic () Quiz 1. Form B Solutions are in RED COLOR رقم التسجيل : الاسم : الشعبة: رقم ====================================================================== Instructions: Time 15 minutes. Closed books and notes. No calculators. No questions are allowed. ======================================================================= Q1. Convert (295) 10 to binary. (8 points) 295 147 1 147 73 1 73 36 1 36 18 0 18 9 0 9 4 1 4 2 0 2 1 0 1 0 1 (295) 10 = 100100111 OR: 295 256 = 39 39 32 = 7 7 4 = 3 3 2 = 1 (295) 10 = 256+32+4 + 2 + 1 = 2 8 + 2 5 + 2 2 + 2 1 + 2 0 = 100100111 Q2. Evaluate the following binary addition. (4 points) Carry 1 1 1 1 0 1 1 0 1 0 0 1 1 1 0 0 0 1 0 0 1 0 0 1 0 Q3. Analyze the following circuit to find Boolean Functions F(X,Y,Z) are being implemented. Then simplify your answer if possible. (16 points) (6 points for the function and 10 points for simplification) F(X,Y,Z) = X(Y + Z)(X+Y+Z ) Before simplification = X [Y (X+Y+Z ) + Z(X+Y+Z )] = X [XY +0+Y Z + XZ+YZ+0] = X [XY +Y Z + XZ+YZ] = X.XY +XY Z +X.XZ+XYZ = XY +XY Z + XZ+XYZ = XY (1+Z ) + XZ(1+Y) = XY + XZ F(X,Y,Z) = XY + XZ After simplification
Q4. Convert (B32) 16 to binary. (4 points) (B32) 16 = 1011 0011 0010 Q5. From the truth table, write the function F(X,Y,Z) in the SOP terms then draw circuit implementing the followin function using AND, OR, and NOT gates without simplification. (18 points) (6 points for the function and 12 points for the circuit) X Y Z F minterm 0 0 0 0 0 0 1 0 0 1 0 1 X YZ 0 1 1 1 X YZ 1 0 0 0 1 0 1 1 XY Z 1 1 0 0 1 1 1 1 XYZ F(X,Y,Z) = X YZ + X YZ + XY Z + XYZ 4